Đáp án:
a. 0
b. 0
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{1}{{\sqrt {7 - 2\sqrt 6 } + 1}} - \dfrac{1}{{\sqrt {7 + 2\sqrt 6 } - 1}}\\
= \dfrac{1}{{\sqrt {\left( {6 - 2.\sqrt 6 .1 + 1} \right)} + 1}} - \dfrac{1}{{\sqrt {\left( {6 + 2.\sqrt 6 .1 + 1} \right)} - 1}}\\
= \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} + 1}} - \dfrac{1}{{\sqrt {{{\left( {\sqrt 6 + 1} \right)}^2}} - 1}}\\
= \dfrac{1}{{\sqrt 6 - 1 + 1}} - \dfrac{1}{{\sqrt 6 + 1 - 1}} = \dfrac{1}{{\sqrt 6 }} - \dfrac{1}{{\sqrt 6 }} = 0\\
b.\dfrac{2}{{\sqrt {8 - 2\sqrt {15} } }} - \dfrac{1}{{\sqrt {5 - 2\sqrt 6 } }} - \dfrac{3}{{\sqrt {7 + 2\sqrt {10} } }}\\
= \dfrac{2}{{\sqrt {5 - 2\sqrt 5 .\sqrt 3 + 3} }} - \dfrac{1}{{\sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} }} - \dfrac{3}{{\sqrt {5 + 2\sqrt 5 .\sqrt 2 + 2} }}\\
= \dfrac{2}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \dfrac{1}{{\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} }} - \dfrac{3}{{\sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} }}\\
= \dfrac{2}{{\sqrt 5 - \sqrt 3 }} - \dfrac{1}{{\sqrt 3 - \sqrt 2 }} - \dfrac{3}{{\sqrt 5 + \sqrt 2 }}\\
= \dfrac{{2\sqrt 5 + 2\sqrt 3 }}{{5 - 3}} - \dfrac{{\sqrt 3 + \sqrt 2 }}{{3 - 2}} - \dfrac{{3\sqrt 5 - 3\sqrt 2 }}{{5 - 2}}\\
= \dfrac{{2\sqrt 5 + 2\sqrt 3 }}{2} - \dfrac{{\sqrt 3 + \sqrt 2 }}{1} - \dfrac{{3\sqrt 5 - 3\sqrt 2 }}{3}\\
= \sqrt 5 + \sqrt 3 - \sqrt 3 - \sqrt 2 - \sqrt 5 + \sqrt 2 \\
= 0
\end{array}\)
